Write the least positive integral value of `n` for which `((1+i)/(1-i))^n` is real.

Options

A. 0
B. 2
C. 4
D. None of these

Answer

To find the least positive integral value of \( n \) for which \( \left( \frac{1+i}{1-i} \right)^n \) is real, we can follow these steps:

### Step 1: Rationalize the expression
We start with the expression:
\[ \frac{1+i}{1-i} \]
To rationalize this, we multiply the numerator and denominator by the conjugate of the denominator, which is \( 1+i \):
\[ \frac{(1+i)(1+i)}{(1-i)(1+i)} \]

### Step 2: Simplify the denominator
Using the difference of squares formula \( a^2 - b^2 \), we simplify the denominator:
\[ (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2 \]

### Step 3: Simplify the numerator
Now, we simplify the numerator:
\[ (1+i)(1+i) = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i \]

### Step 4: Combine results
Now we can rewrite the expression:
\[ \frac{1+i}{1-i} = \frac{2i}{2} = i \]

### Step 5: Raise to the power of \( n \)
Now we have:
\[ \left( \frac{1+i}{1-i} \right)^n = i^n \]

### Step 6: Determine when \( i^n \) is real
The powers of \( i \) cycle through four values:
- \( i^1 = i \) (not real)
- \( i^2 = -1 \) (real)
- \( i^3 = -i \) (not real)
- \( i^4 = 1 \) (real)

Thus, \( i^n \) is real when \( n \) is even. The smallest positive even integer is \( n = 2 \).

### Conclusion
The least positive integral value of \( n \) for which \( \left( \frac{1+i}{1-i} \right)^n \) is real is:
\[ \boxed{2} \]